Question

A solution contains 0.1 M $H_{2}S$ and 0.3 M HCl. Calculate the conc. of $S^{2-}$ and $H{S}^{-}$ ions in solution, Given $K_{a_1}$ and $K_{a_2}$ for $H_{2}S$ are ${10}^{-7}$ and $1.3\times {10}^{-13}$ respectively.

• A. $1.43\times {10}^{-20}M,3.3\times {10}^{-8}M$
• B. $1.43\times {10}^{-10}M,3.3\times {10}^{-4}M$
• C. $2.83\times {10}^{-19}M,3.3\times {10}^{-8}M$
• D. $1.43\times {10}^{-19}M,1.65\times {10}^{-8}M$

Answer 1

The correct option is A $1.43\times {10}^{-20}M,3.3\times {10}^{-8}M$

$H_{2}S\rightleftharpoons H^{+}+H{S}^{-};K_{a_1}={10}^{-7}$
$H{S}^{-}\rightleftharpoons H^{+}+S^{2-};K_{a_2}=1.3\times {10}^{-13}$
$HCl\to H^{+}+C{l}^{-}$
Due to common ion effect exerted by $H^{+}$ of HCl, the dissociations of $H_{2}S$ are suppressed and the $\left[H^{+}\right]$ in solution is mainly due to HCl.
$\therefore K_{a_1}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}$
$\Rightarrow {10}^{-7}=\frac{\left[0.3\right]\left[H{S}^{-}\right]}{\left[0.1\right]}[\because [H^{+}]$ from HCl = 0.3 and $H_{2}S$ would have dissociated negligibly]
$\therefore \left[H{S}^{-}\right]=\frac{{10}^{-7}\times 0.1}{0.3}=3.3\times {10}^{-8}M$
Further $K_{a_2}=\frac{\left[H^{+}\right]\left[S^{2-}\right]}{\left[H{S}^{-}\right]}\Rightarrow 1.3\times {10}^{-13}=\frac{\left[0.3\right]\left[S^{2-}\right]}{3.3\times {10}^{-8}}$
$\therefore \left[S^{2-}\right]=\frac{1.3\times {10}^{-13}\times 3.3\times {10}^{-8}}{0.3}=1.43\times {10}^{-20}M$